# Evaluating $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$

I am trying to evaluate $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$. Factoring and using partial fraction decomposition, I have found that the indefinite integral is: $$\ln{|x-5|}-\ln{|x-3|} + C$$

But when evaluating the improper integral, I don't know how to deal with the difference between two diverging limits, or even whether or not it even converges. My intuition tells me that since the denominator of the original integral is a second degree polynomial, it converges, but I'm not entirely sure.

$$\lim_{b\to\infty}{(\ln{(b-5)}-\ln(b-3))}$$

How do you evaluate this limit?

• You can complete the square. – Mhenni Benghorbal May 20 '13 at 1:05

Combine and you have $$\log(b -5) - \log(b - 3) = \log\left({b - 5 \over b - 3}\right).$$ Exploit the continuity of the log function at 1 to finish. This is only useful if your start point for integrating is beyond the nasty places $x = 3$ and $x = 5$. Otherwise integrating across these points will render the integral nonexistent.
Use that $$\log x-\log y=\log\dfrac xy$$
It doesn't converge in the traditional sense, as it is unbounded at $x = 5$ and $x = 3$ (and is unbounded in such a way that the integral does not converge). It does have a finite Cauchy Principal Value though.